Local minimum /Maximum

Question

Find a cubic function, in the form below, that has a local maximum value of 4 at -4 and a local minimum value of 0 at 3.

f (x) = ax³ + bx² + cx + d

Answer 1

f'(x) = 3ax² + 2bx + c = 0. It has 2 solutions - 4 and 0;

so, c = 0; 48a - 8b = 0 or 6a - b = 0.

Now f(0) = d = 3; f(-4) = - 64a + 16b + 3 = 4; - 64a + 16b = 1; 96a - 16b = 0; 32a = 1; a = 1/32; b = 3/16.

f(x) = 1/32x³ + 3/16x² + 3;

f'(x) = 3/32x² + 3/8x; f''(x) = 3/16x + 3/8; f''(3) = 3/4 > 0, so, at 3 we have local minimum

f''(- 4) = - 3/4 + 3/8 = - 3/8 < 0.\; so, at x = - 4 we have local maximum.

Answer: f(x) = 1/32x³ + 3/16x² + 3

Answer 2

Hint:

f (x) = ax³ + bx² + cx + d

f'(x) =3ax² + 2bx +c

f(-4) = -64a + 16b -4c +d =4

f(3) = 27a + 9b + 3c + d = 0

Since local maximum and minimum, the first derivative will be zero:

f '(-4) = 48a -8b + c = 0

f '(3) = 27a+6b + c =0

There are four equations and four unknowns, solve for a, b, c and d.